There are $n$ components in parallel and at least $m$ of them are required to work (in order for the system to work); so we can have $n-m$ or less failures. The $n$ components are made up of $a$ of type $\alpha$ and $b$ of type $\beta$ such that $a+b=n$. The Let $z$ be the amount of failures so it ranges from $0$ to $n-m$, the failures are made up $x$ of type $\alpha$ and $y$ of type $\beta$ so that $x+y=z$. The reliability of a type $\alpha$ component is $R_\alpha$ and for type $\beta$ it is $R_\beta$, each failure is independent (i.e. any $\alpha$ is identical to any other).
I thought that we could use a multinomial coefficient here; so there are $n$ components and we wish for each one to fall into one of four categories:
- a failed type $\alpha$; probability $1-R_\alpha$
- a working type $\alpha$; probability $R_\alpha$
- a failed type $\beta$; probability $1-R_\beta$
- a working type $\beta$; probability $R_\beta$.
We want there to be $x$ falling into the first category, $a-x$ in the second etc. Hence, I thought we could use the following formula for the chance of $z$ failures:
${n\choose x,a-x,y,b-y}(1-R_\alpha)^xR_\alpha^{a-x}(1-R_\beta)^yR_\beta^{b-y}.$
We would then sum over $z$ from $0$ to $n-m$ as $z$ takes each value mutually exclusively (either there is 1 failure or 2 or etc.). I think we might also have to consider how to factor in all the different ways in which $x$ and $y$ can make up $z$ (see the options below) but I'm really not sure.
The issue here is that the value of the multinomial coefficients are far far too large (in the millions!) and obviously the reliability of the system must be between 0 and 1, so something is pretty wrong but I am having trouble seeing the flaw in my reasoning.
I've tried thinking about it in a different way.
Take the case where there is one failure ($z=1$), then the mutually exclusive options for failing components are
- $\alpha$; this can happen ${1\choose 0}=1$ way; probability is ${a\choose 1}(1-R_\alpha)^1R_\alpha^{a-1}{b\choose 0}(1-R_\beta)^0R_\beta^{b}$ (${a\choose 1}=a$ because there are $a$ possible ways to choose the 1 $\alpha$ that will fail from the $a$ of type $\alpha$ etc.)
- $\beta$; this can happen ${1\choose 1}=1$ way; probability is ${a\choose 0}(1-R_\alpha)^0R_\alpha^{a}{b\choose 1}(1-R_\beta)^1R_\beta^{b-1}$
Now if $z=2$
- $\alpha\alpha$; ${2\choose 0}=1$ ways; probability ${a\choose 2}(1-R_\alpha)^2R_\alpha^{a-2}{b\choose 0}(1-R_\beta)^0R_\beta^{b}$
- $\alpha\beta$; ${2\choose 1}=2$ ways*; probability ${a\choose 1}(1-R_\alpha)^1R_\alpha^{a-1}{b\choose 1}(1-R_\beta)^1R_\beta^{b-1}$
- $\beta\beta$; ${2\choose 2}=1$ ways; probability ${a\choose 0}(1-R_\alpha)^0R_\alpha^{a}{b\choose 2}(1-R_\beta)^2R_\beta^{b-2}$
The reliability from the $\alpha\beta$ combination is the same as for $\beta\alpha$, however if we say an $\alpha$ component fails, then this rules out the possibility of a $\beta\beta$ combination, but it is still possible to have an $\alpha\beta$ reliability, vice versa if we swap $\alpha$ and $\beta$. In this sense we are twice as likely to have an $\alpha\beta$ combination compared to an $\alpha\alpha$ or $\beta\beta$ combination. So it seems like we should multiply the reliability for this option by a factor of 2. However, I am not sure if this notion of considering the orders contradicts the failures all being independent and each $\alpha$ and each $\beta$ being identical. Are we definitely considering unordered sample groups and what exactly does this mean?
If the latter method is more correct then I have the formula
$\sum_{z=0}^{n-m}\sum_{r=0}^{z}{z\choose r}{a\choose z-r}(1-R_\alpha)^{z-r}R_\alpha^{a-(z-r)}{b\choose r}(1-R_\beta)^rR_\beta^{b-r}.$
Here the failures are made up of $(z-r)+r=z$ and the remainder of the components are accounted for with the $(a-(z-r))+(b-r)=a+b-z$ which then combine together as all $z+(a+b-z)=a+b=n$ of the components.
I'm sorry this is so long but I could really use someone to point out flaws in my reasoning and suggest a way forward...
Thanks!
